Chapter 10.4, Problem 15E

To determine

Tofind:The radius of convergence of given power series.

Answer to Problem 15E

The radius of convergence of the series ${\displaystyle \sum _{n=0}^{\infty }\frac{n{\left(x+3\right)}^n}{5^n}}$ is $R=5$ .

Explanation of Solution

Given information:

The given power series is ${\displaystyle \sum _{n=0}^{\infty }\frac{n{\left(x+3\right)}^n}{5^n}}$ .

Methodused:

Theorem:Let ${\displaystyle \sum _{n=0}^{\infty }{c}_n}{\left(x-a\right)}^n$ be a given series and L be a positive number such that $\underset{n\to \infty }{\lim }\frac{c_{n+1}}{c_n}=L$ , then the radius of convergence of series is $R=\frac{1}{L}$ .

Then there exists a number R such that if $|x-a|<R$ the series converges, if $|x-a|>R$ series diverges. That number R is called the radius of convergence of the series ${\displaystyle \sum _{n=0}^{\infty }{c}_n}{\left(x-a\right)}^n$ .

Calculations:

The given series is ${\displaystyle \sum _{n=0}^{\infty }\frac{n{\left(x+3\right)}^n}{5^n}}$ .

Thus, we have ${\displaystyle \sum _{n=1}^{\infty }{c}_n}{\left(x-a\right)}^n={\displaystyle \sum _{n=0}^{\infty }\frac{n{\left(x+3\right)}^n}{5^n}}$

  $\Rightarrow c_n=\frac{n}{5^n}$

Now, the limit L is $L=\underset{n\to \infty }{\lim }\frac{c_{n+1}}{c_n}=\underset{n\to \infty }{\lim }\frac{\frac{n+1}{5^{n+1}}}{\frac{n}{5^n}}$

  $\Rightarrow L=\underset{n\to \infty }{\lim }\frac{n+1}{n}\times \frac{5^n}{5^{n+1}}$

  $\Rightarrow L=\underset{n\to \infty }{\lim }\left(1+\frac{1}{n}\right)\times \frac{1}{5}$

  $\Rightarrow L=\underset{n\to \infty }{\lim }\left(1+\frac{1}{n}\right)\times \frac{1}{5}$

  $\Rightarrow L=\left(1+0\right)\times \frac{1}{5}$$\because \underset{n\to \infty }{\lim }\frac{a}{n}=0$

  $\Rightarrow L=\frac{1}{5}$

  $\Rightarrow R=\frac{1}{L}=5$

Thus, the radius of convergence of the series ${\displaystyle \sum _{n=0}^{\infty }\frac{n{\left(x+3\right)}^n}{5^n}}$ is $R=5$ .

Conclusion:

The radius of convergence of the series ${\displaystyle \sum _{n=0}^{\infty }\frac{n{\left(x+3\right)}^n}{5^n}}$ is $R=5$ .